CHEM 2410: Exam 4: 8.1-8.4 Addition Reactions, Markovnikov

Addition

2 molecules combining to form 1 product, often adding across a double C—C bond to saturate it

Elimination

1 molecule splitting into fragment molecules, often an alkene, as well as the products of two eliminated attachments (original leaver like a halogen, as well as an H)

Substitution

When one part of a molecule is replaced by another fragment in a molecule 

Which direction do alkene reactions typically move and why?

They go from being unsaturated (double bonds;sigma+pi) to saturated (single bonds) because sigma/single bonds are more stable than pi bonds, and we want to achieve the most stable state, which is moving an alkene to an alkane

Why are sigma bonds more stable than pi bonds?

Electrons in sigma bonds are held tightly between the two bonding molecules, so they are unable to move/restricted

Electrons in the pi bond are above and below the molecules, meaning they are not held as tightly and are instead delocalized, which allow them to be more likely to react with nucleophiles.

What is the General Process of Addition to Alkenes?

Strong electrophiles attract loose electrons from pi bonds of an alkene, and forms sigma bond with one of the Carbons from the double bond, while the other gains a carbocation, which is a strong electrophile. Now, this electrophile acts with a weak nucleophile to from another sigma bond

What is the Sequential Process of Addition to Alkenes?

Step 1: attack of pi bond on electrophile, so a Carbon on the double bond gains that electrophile as a single bond, the double bond turns into a single bond, and the other Carbon on the double bond gains a carbocation 

Step 2: The carbocation is a strong, positive electrophile, which allows a weak nucleophile to come in and bond to it, eliminating the carbocation

How do you approach Markonikov’s Rule in Addition Reactions?

Step 1: Protonate the double bond (negative double bond to positive Hydrogen on electrophile), but it depends which Carbon you add it to. You add the Hydrogen to the least substituted intermediate, so that the carbocation can be on the most substituted molecule.

Step 2: Add on the nucleophile to the carbocation.

What is the Ionic Addition of H—X to an Alkene Mechanism?

Step 1: Protonate pi bond to a Hydrogen forming a C—H bond on the least substituted Carbon, and a carbocation on the more substituted Carbon

Step 2: Extra nucleophilic halide ion will now bond to the carbocation to give an addition product.

Hammond’s Postulate 

a transition state will have a structure that is similar to the species (reactant, intermediate, or product) it is closest to in energy

for exothermic reactions, transition states will most resemble the reactants, and vice versa 

Regioselective

1 of the 2 orientations in addition preferred over another 

Markonikov Rule

Must add the proton to the least substituted side, and the non-H atom to the more substituted side

You must add the electrophile in a way so you can generate the more stable intermediate 

What triggers the Anti-Markonikov Addition Reaction?

Peroxides R—O—O—R form free radicals mechanisms

R—O—O—R have a weak bond to each other so they break easily, leaving them each with one radical R—O.

What are radicals considered?

Electrophiles since they do not have a complete octet so they are missing an electron so they are relatively more positive

What does this property do that causes radicals to trigger an anti-Markonikov radical reaction?

Since they are electrophiles, this causes double bonds to attach to them to complete the octet, leaving the other electrons as radicals on the other carbon

Explain the Mechanism of Radical Addition to HBr of an Alkane

Initiation: Formation of Radicals

• R—O—O—R —Heat—> R—O. + .O—R

Propagation: Radical Reacts to Generate Another Radical from HBr

• R—O. + H—Br —> R—O—H + Br-

Bromide Radical Adds to Double Bond to Generate Alkyl Radical on More Substituted Carbon

• Br- + C=C —> Br—C—C.

Alkyl Radical Abstracts H Atom from HBr to Generate a Product + Another Br Radical 

• Br—C—C. + H—Br —> Br—C—H + Br.

Chain Reaction Continues, Until 2 Radicals Combine and Terminate Reaction

How do Radicals Create Anti-Markonivkov Products

Bromine attaches to the Carbon first, meaning the other side, which is more substituted, gets the radical. That means instead of the Bromine getting to go on the more substituted side, the Hydrogen goes, as it is second.

How do radical reactions differ from ionic addition reactions?

In ionic addition, H is the electrophile and reacts with the alkene first, but in radical addition, Br. is the electrophile and reacts with the alkene first 

Peroxide Effect

Reversed in orientation from adding the bigger atom on the more substituted side, forming an anti-markovnikov product due to peroxides, ONLY for addition! 

Hydration

Adding H2O to an alkene, so that H is added to one C, and OH is added to another C, and has a mechanism exactly reversed from dehydration

Principle of Microscopic Reversibility

Hydration mechanism is exactly the reverse of the dehydration mechanism

Explain the Acid Catalyzed Hydration Mechanism on an Alkene

Step 1: Protonate the double bond to create a carbocation on one C and a C—H bond on another C, as well as neutral water
Alkene and H3O+—> H—C—C+

Step 2: Nucleophilic Attack by H2O Gives Protonated Alcohol, where the molecule gets attacked by a lone pair of the neutral water molecule which acts as a nucleophile, so that the entire water molecule attaches to the electrophilic carbocation

H—C—C—HOH

Step 3: Deprotonate H2O Gives H3O+, have another water molecule attack a Hydrogen on the H2O attached to the molecule, to form a molecule with bonds of H and OH, as well as an acidic H3O+

H—C—C—HOH + H2O —> H—C—C—OH + H3O+

Basic Characteristics of Hydration

Markovnikov Addition, H2O attacks CC to attach itself to CC, another H2O attacks to deprotonate the H2O attached to the molecule to form H3O+